One gram of charcoal adsorbs 100 ml of 0.5 M `CH_3COOH` to form a monolayer and thereby the molarity of acidic acid is reduced to `0.49M`. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid (surface area of charcoal is `3.01 xx 10^(2) m^2//g)`.

Explanation

Number of moles of acetic acid before the adsorption = `0.5 xx 100/1000 = 0.05`
Number of moles of acetic after the adsorption = `0.49 xx 100/1000 = 0.049`
Number of moles of acetic acid adsorbed = `0.05 - 0.049 = 0.001`
Number of molecules of acetic acid adsorbed = `0.001 xx 6.023 xx 10^(23) = 6.023 xx 10^(23)`
Surface area of the charcoal occupied by each acetic acid molecule = `(3.01 xx 10^2)/(6.023 xx 10^(20)) = 5 xx 10^(-19) m^(2)` .